Conics query (1 Viewer)

[Avengelion]

Hi this was a question in 2003 HSC question 4 b) iii), I think it would be best if a diagram is drawn. You have a hyperbola x2/a2 - y2/b2 = 1 as usual, and the usual tangent xsec@/a - ytan@/b = 1. The tangent cuts the asymptotes at A and B. Prove that the area of the triangle OAB = ab.

Ok, there are only two methods in find triangle areas:

1) perpendicular distance * side/2
2) Sine method

I looked at the solutions and it used 2), where they let angle BOA = 2X, where X is equal to tan^-1 (b/a). If the tangent was perpendicular to the x-axis, X would be easy to prove, but how can u prove that X = tan^-1(b/a) if the tangent varies?

Help is very appreciated.

 

[Mattamz]

the value of x is independent of the tangent, as it is angle between the aysmtopes and the x-axis(fig1). thus by considering a triangle (fig2) x = tan-1(b/a).

 

[Yip]

X represents the angle between the x-axis and the asymptote, not the angle between the tangent and the x-axis. Its quite clear if u draw the diagram.

For your interest, there is actually a third method to do it, but it involves university mathematics, more specific, it involves the use of determinants.

In this question, the coordinates of A, which represents the intersection with the negative gradient asymptote are [a/(sec@+tan@),-b/(sec@+tan@)], coordinates of B are [a/(sec@-tan@),-b/(sec@-tan@)]. The coordinates of O are (0,0).

The area of a triangle in a 2-dimensional plane, given its 3 vertices, can be expressed as:

|x1 y1 1|
|x2 y2 1|*0.5 =0.5*[x1(y2-y3)-y1(x2-x3)+1(x2y3-x3y2)]
|x3 y3 1|
The expression above is called a determinant. Using this formula for the vertices given above, the area ends up to be 0.5*2ab=ab. Note that the above formula gives the signed area, i.e. depending on what coordinates u choose to be (x1,y1), (x2,y2), (x3,y3), u may get a negative result. To get a positive result everytime, u must cycle through the points in an anticlockwise direction, i.e. if u let (0,0) be (x1,y1), then u must make A to be (x2,y2) and B to be (x3,y3).

Anyway, that method is a lot faster than the other 2 methods, instead of a page of messy algebra, u get the answer in 2 lines. However, try not to use it in exams unless u cant see any other way/u are running out of time. Markers tend to get a little annoyed if people start diverging from the syllabus.

 

[Avengelion]

OMG OMG OMG!!!

I can't believe I couldn't see it... I guess too many past papers really stuffs up the head. Thanks alot.

O btw Yip, that's a very interesting method there, its like, you could do the question in less than a minute and its true that I got all a load of long algebra before I could get the answer.

Anyway, thanks alot again.

 

[kony]

determinant method is probably allowed.

you could always memorise its proof, which is quite short anyway.